Ans :$\textbf{2018}$ of course

Let $t=\gcd(a,b)$ i.e $(a,b)=(tx,ty)$ where $\gcd(x,y)=1$. $$\frac{ab}{2b^2-a}=\frac{a^2+4ab+4b^2}{16b^2}$$$$16b^3a=(a^2+4ab+4b^2)(2b^2-a)$$$$16ty^3x=(x^2+4xy+4y^2)(2ty^2-x)$$Firstly notice $y\mid (x^2+4xy+4y^2)(2ty^2-x)$. Moreover $\gcd(y,x^2+4xy+4y^2)=\gcd(y,x^2)=1$ and $\gcd(y,2ty^2-x)=\gcd(y,-x)=1$ so it follows $y=1$. Indeed $$16tx=(x^2+4x+4)(2t-x)$$Taking modulo $2$ we get $x=2k$ so $$4tk=(k^2+2k+1)(t-k)$$Easy to see $k,t$ are both odd, so since $\gcd(k,k^2+2k+1)=1$ we have $k\mid t-k$ i.e $t=km$ i.e $$4km=(k^2+2k+1)(m-1)$$Now since $\gcd(m,m-1)=1$ and $\gcd(k,k^2+2k+1)=1$ we have $m\mid k^2+2k+1$ and $k\mid m-1$ so $m=uk+1$ and we have $$uk+1\mid k^2+2k+1$$Set $k^2-k(un-2)+1-n=0$ if $k_0\in\mathbb N$ is a solution let $p$ be the other, so $k_0+p=un-2$ and $k_0p=1-n$. So we can conclude $p\in\mathbb Z^-\cup\{0\}$. If $p\neq 0$ so $n\neq 1$ we have $k_0> un-2$ but also $k_0\le n+1$ so $n+1>un-2$ this is $n+3>un$. If $n>2$ this implies $u=1$. But if $u=1$ we get $m=k+1$ and $4(k+1)=(k+1)^2$ implies $k=3$ for which we get $x=6$ and $t=12$ so $(a,b)=(72,12)$. Now if $n\le 2$ we can have $u=1$ or $u=2$ but $u=2$ implies $2k+1\mid k^2+2k+1$ implies $2k+1\mid k^2$ so $k=0$ contradiction. Now the rest is consider when $p=0$. If $p=0$ we have $n=1$ i.e $m=uk+1=k^2+2k+1$ so $4k(k+1)^2=(k+1)^2((k+1)^2-1)$ give us $4k+1=(k+1)^2$ which is $2k=k^2$ so $k=2$ which give us $(a,b)=(72,18)$. So the only solutions to the diophantine $$\sqrt{\dfrac{ab}{2b^2-a}}=\dfrac{a+2b}{4b}$$are $(a,b)=(72,12),(72,18)$ @below: I noticed $(72,18)$ (my first edit) but I just wrote was $un-2>0$, moreover I had a line $2k=k^2$ a clearly contradiction LOL

Darn somehow I kept thinking that $10\cdot 200+8=2018$ nooooo. Anyway, beautiful solution @above! Here is a weird logic way to get the answer. So, note that the year is $2018$. Then observe that $|10(a-5)(b-15)|+8$ looks very strange, as if $2018$ is begging to be the answer due to the absolute value and $+8$. If so, then $(a-5)(b-15) = \pm 201$ and note that $201=\pm 3\cdot \pm 67$. Guess and check gives $(72,18)$.

hsiangshen wrote: Let $a,b$ be positive integers satisfying $$\sqrt{\dfrac{ab}{2b^2-a}}=\dfrac{a+2b}{4b}$$. Find $|10(a-5)(b-15)|+8$. $\sqrt{\frac{ab}{2b^2-a}} =\frac{a+2b}{4b}$ By squaring both the sides we get $16b^3a=(2b^2-a)(a+2b)^2=2b^2a^2+8b^4+8ab^3 -a(a+2b)^2$ $\implies 2b^2(a-2b)^2=a(a+2b)^2 \implies 2b|a$ So let $a=2bx$ then we have $b(x-1)^2=x(x+1)^2\implies x-1|x+1$ 1-) $x=2$ hence $a=4b$ By solving we get $(a, b)=(72, 18)$ 2-) $x=3$ hence $a=6b$ By solving we get $(a, b)=(72, 12)$ @below *Edited ;-)

Above $x-1|x+1\Rightarrow x-1|2$ You forgot $x=3$